Question

what is the equivalent mass of KMNO4 in acidic and alkaline medium?

Answer 1

In Acidic medium, This Mn+7 goes to Mn+2 state and hence there is a net gain of 5 electrons.
Now, equivalent weight = [molar mass] /[number of electrons gained or lost]
so, eq wt = 158/5 = 31.6 g.

Now, for alkaline medium, there are two possibilities
1) faintly alkaline/neutral:
Mn+7 changes into Mn+4 therefore, gain of 3 electrons
Hence, eq wt = 158/3 = 52.67g.
2) highly alkaline:
Mn+7 changes into Mn+6 therefore, gain of 1 electron
Hence, eq wt = 158/1 = 158g

In many cases though if it's written alkaline, it mostly means the neutral one; for the highly alkaline thing, it'll especially mention it.

Answer 2

Answer: Equivalent mass of $KMNO_4$ in acidic and alkaline medium is 31.6 and 52.7 respectively.

Explanation:

${\text {Equivalent mass}}=\frac{\text {Molecular mass}}{\text {number of electrons involved}}$

In acidic medium the number of electrons involved are 5.

$MnO_4+8H^++5e^{-1}\rightarrow Mn^{2+}+4H_2O$

Thus ${\text {Equivalent mass of}KMNO_4=\frac{158}{5}=31.6$

In basic medium the number of electrons involved are 3.

$MnO_4+2H_2O+3e^{-1}\rightarrow MnO_2+4OH^-$

Thus ${\text {Equivalent mass of}KMNO_4=\frac{158}{3}=52.7$